已知函数f(x)=cos(π/6-2x)+cos(2x+π/6)+sin(2x+π/3)-sin(π/3-2x).

题目

已知函数f(x)=cos(π/6-2x)+cos(2x+π/6)+sin(2x+π/3)-sin(π/3-2x).
⑴:求函数f(x)的最小正周期.
⑵:在锐角△ABC中,角ABC所对的边分别为abc,且f(A)=1,a=1,试求△ABC的面积S的最大值

答案

f(x)=cos(π/6-2x)+cos(2x+π/6)+sin(2x+π/3)-sin(π/3-2x)=sin(π/2-(π/6-2x))+cos(2x+π/6)+sin(2x+π/3)-cos(π/2-(π/3-2x))=sin(2x+π/3)+cos(2x+π/6)+sin(2x+π/3)-cos(2x+π/6)=2sin(2x+π/3)最小正周期是...